Re: [理工] [機率]-probability density function

[mokiya1 (無言)]

※ 引述《swatch0811 (............)》之銘言：
:         因為小第我沒修過機率，自己看書後，
:         又覺得寫出來的答案很不確定，煩請會
:         的人，幫我解答一下(希望可以大R回復)
:         謝謝。
: -----------------------------------------------------
:         Q1: Let random variables Y = X^2 .Find the probability density
:             function and expectation of Y for the following cases.
:             1. X takes the value -2, -1, 0, 1, 2, 3 with
:                equal probability 1/6.
:             2. X is uniformly distributed in the interval (0,1).
:         Q2: There are two random variables X 屬於 {0, 1, 2} and
:                 ┌ 1  ,  if X = 0
:             Y = │
:                 └ 0  ,  if X = 1,2
:             If P(X=0) = P(X=1) = P(X=2) = 1/3, and P(Y=0) = P(Y=1) = 1/2,
:             then are X and Y orthogonal? uncorrelated? independent?
Q1:
1.
x=-2,x=2, y=4 p(y=4)=1/3
x=-1,x=1, y=1 p(y=1)=1/3
x=0,y=0       p(y=0)=1/6
x=3,y=9       p(y=9)=1/6

fy(y)= 1/6 ,y=0
      1/3 ,y=1
      1/3 ,y=4
      1/6 ,y=9
2.X~U(0,1)
  Y=X^2 , X=根號Y
fx(x)=1
fy(y)=fx(y)*dx/dy=1*1/2*1/根號y=1/2根號y,0<y<1
Q2:
1.orthogonal=>E[XY]=0
E[XY]=E[0]=0->orthogonal
2.uncorrelated=>COV[X,Y]=0
COV[X,Y]=E[XY]-E[X]E[Y]
E[X]=0*1/3+1*1/3+2*1/3=1
E[Y]=0*1/2+1*1/2=1/2
COV[X,Y]=-1/2
所以correlated
3.independent=>fxy=fx*fy
f(x=0,y=0)=0
f(x=0)=1/3
f(y=0)=1/2
f(x=0)*f(y=0)=1/6不等於f(x=0,y=0)
所以dependent

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◆ From: 114.41.117.87
推 swatch0811:謝謝^^  我再試試看
推 swatch0811:我想問E[XY]=E[0+0] 為何是寫E["XY"]卻又用加法?
→ swatch0811:還有COV是什麼意思^^"  謝謝
※ 編輯: mokiya1         來自: 114.41.117.87        (03/22 13:46)
→ mokiya1:打錯...
→ mokiya1:cov是共變異數,定義我已經寫得很詳細了
推 swatch0811:懂了@@